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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2020 · Mathematics · Indefinite Integration

If p(x)=(4 e)^ 2 x , then what is ∫ p(x) d x equal to?

If p(x)=(4e)2xp(x)=(4 e)^{2 x}, then what is ∫p(x)dx\int p(x) d x equal to?

Options

  1. Ap(x)1+2ln⁡2+c\frac{p(x)}{1+2 \ln 2}+c
  2. Bp(x)2(1+2ln⁡2)+c\frac{p(x)}{2(1+2 \ln 2)}+c
  3. C2p(x)1+ln⁡4+c\frac{2 p(x)}{1+\ln 4}+c
  4. Dp(x)1+ln⁡2+c\frac{p(x)}{1+\ln 2}+c

Correct answer

B. p(x)2(1+2ln⁡2)+c\frac{p(x)}{2(1+2 \ln 2)}+c

Step-by-step solution

∫P(x)⋅dx=∫(4⋅e)2x⋅dx=∫e2x(ln⁡e+ln⁡4)⋅dx=∫e2x(1+ln⁡4)⋅dx=e2x(1+ln⁡4)2(1+ln⁡4)+C\int P(x) \cdot d x=\int(4 \cdot e)^{2 x} \cdot d x=\int e^{2 x(\ln e+\ln 4)} \cdot d x=\int e^{2 x(1+\ln 4)} \cdot d x=\frac{e^{2 x(1+\ln 4)}}{2(1+\ln 4)}+C \quad where C=C= constant =(4e)2x2(1+2ln⁡2)+C∴∫P(x)⋅dx=P(x)2(1+2ln⁡(2))+C.\begin{aligned} & =\frac{(4 e)^{2 x}}{2(1+2 \ln 2)}+C \\ & \therefore \int P(x) \cdot d x=\frac{P(x)}{2(1+2 \ln (2))}+C .\end{aligned}
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